For many of these problems use KEi + Ui
= KEf + Uf + HEAT or
USE KEi + Ui + WF
= KEf + Uf + HEAT, where U is the
potential energy due to the conservative force (
gravitational and spring forces in this case), KE is the
kinetic energy and WF is the work done by
the external force (done by "Frankie" as we
discussed in class. )
NOTE: 9, 46 and 63 mix ENERGY CONSERVATION (Ch. 7)and
CIRCULAR DYNAMICS (Ch. 5), a special class of problems
you should study ahead of Test 2.
ANOTHER NOTE: #50 mixes ENERGY CONSERVATION (Ch. 7) and PROJECTILE
MOTION (CH. 3)

9.* KEi + Ui = KEf + Uf
+ HEAT .
(a) The angle between the normal force and motion in this case is
90 degrees, so what is the work by that force? The work by
gravity is positive since the motion is from high to low; note the
gravitational work only depends on the vertical displacement.
(b) ENERGY CONSERVATION : KEi + Ui =
KEf + Uf + HEAT where HEAT IS
THE NEGATIVE OF THE WORK BY FRICTION. THUS HEAT IS ALWAYS
POSITIVE. You are given the heat.
(c) CIRCULAR DYNAMICS: Critical point. HEAT is the
negative of the work by the friction force, which
may not be constant. The instantaneous friction force
is u*N, where u is the kinetic friction coefficient
and where N is the magnitude of the normal force which may not be
constant as in this case. Clearly from earlier problems in
Chapter 5, N is a maximum at the bottom and is a minimum at the
top. (See worksheet
2, GRP 3, problem #1, THE PROBLEM DEALING WITH
BUMPS AND VALLEYS)
(d) CIRCULAR DYNAMICS: At bottom, the normal
force points up in the pos direction toward the center
of the circle and the weight of magnitude mg points down in the neg
direction. Write:
Sum of forces in radial direction = m*v2/R = pos
- neg and solve for N, the normal force
magnitude, using the speed v you found in part (b)
from energy conservation. See worksheet
2, GRP 3, problem #1. Note N > mg.

(b) If you think the acceleration is constant think twice; the
spring makes things more complex as we have already seen in
Ch. 14. So use: sum of the forces in the y
direction = m*a = pos - neg, where we set the positive y direction
to be up. Thus: : m*a = | kyf' | + fk
- mg, where clearly the spring and friction
forces point upward as the object moves down.

38.* READ section 7.4 and 7.5: If U =
U(x) then Fx = -dU/dx. The x-component of force is the
negative derivative of U; Thus if dU/dx is positive , the
force points in the negative x -direction (Fx <
0) and if dU/dx is negative then force points in positive
x direction (Fx > 0.) When dU/dx = 0 force
is zero and the object is said to be in equilibrium, either
stable or unstable. A CONCAVE UP POTENTIAL ENERGY FUNCTION
MEANS EQUILIBRIUM IS STABLE.

41*. (a) CHAPTER 5 REVIEW: You must prove whether or not
the system is at rest; the emphasis here is the use of Ch. 5
concepts, whereas Ch. 7 is more directly related to part
(b).

If the system is at rest and remains so we have for the
concrete m*a = pos - neg = mg - T = 0 . So you can immediately
find T.

For the Box we have m'*a = 0 = T - fs
' , where fs ' is the force of static friction from the
ground and m' is the Box mass. We also have m''*a =
0, where m'' is the gravel mass. ( i.e. There is no friction
force between the gravel and Box if the system is at rest.)
Compare T and fsmax = us*N, where N is
the magnitude of the normal force on the box, which must be equal
to (m' + m'')*g.

If T < fsmax ', then the system is at rest and remains
so and you can find the friction force of magnitude fs
' on Box.

(b) Before you begin this part, you should check if
the system will move under this new condition. Assume the
system is at rest and will remain so; if you get a
contradiction then the assumption is wrong--- this is
called proof by contradiction, common in math classes. If the
system is at rest and remains so, then fs ' = T < fsmax
' . When the gravel is removed, the maximum friction force
magnitude now is fsmax ' = us*m'*g , now
smaller.

But T = m*g = cement weight
magnitude > fsmax ', contradicting the
assumption. The system won't remain at rest and you can now
use :

KEi + Ui = KEf + Uf
+ HEAT, where KEi = 0 and
Ui = m*g*yi, where yi = 2.00
m. (The lowest point for the cement in this problem is
considered to be the zero of gravitational potential energy; note
we did not include the gravitational potential of the Box
since it does not change. If we included it, it would cancel out
from both sides of the equation.)
HEAT = fk*D , where fk is the friction force
magnitude on Box given by CH. 5 THEORY: f = u*N, where u is
the coefficient of kinetic friction. Clearly the vertical drop
distance of cement equals the horizontal distance moved by
Box.

42.* KEi + Ui = KEf + Uf
+ HEAT , where HEAT = 0.

(a) See
VIRTUAL LAB 2, question 1.
(b) The simplest way to do the problem is to bypass the
speed and kinetic energy just after the Block leaves
the spring and consider only two points in space-time: When the
spring is fully compressed and the block is at rest (i) and when
the block has reached its maximum height at the top
and is momentarily at rest before turning around and sliding back
down (f).
KEi + Ui = KEf + Uf
. You should be able to see which terms are clearly zero. Note:
Let the horizontal level be defined y = 0 and
the zero of gravitational potential energy. Thus we have
initial (i) spring potential energy = final (f) gravitational
potential energy at the top . Once you get the final
potential energy at top find the vertical height H and
use trig to find the distance along the
incline.

43.* The simplest way to do the problem is to bypass the speed and
kinetic energy just after the Block leaves the spring
and consider only two points in space-time: When the spring is fully
compressed and the block is at rest (i) and when the block
has reached its maximum horizontal distance and comes to
rest permanently (f) . KEi + Ui = KEf +
Uf + HEAT , where Ug = 0 at both points
(ground level, y = 0) . You should be able to see which terms are
clearly zero. Thus we have initial (i) spring potential energy =
HEAT , where HEAT = fk*D :

(1/2)*k*d2 = u*mg*D, where u = coefficient of
kinetic friction and d and D are given in diagram.

CIRCULAR DYNAMICS: ALSO AT TOP, the normal force and
the weight of magnitude mg point down in the pos direction
toward the center of the circle. Write:
(II) Sum of forces in radial direction = m*v2/R =
pos - neg = N + mg - 0 = N + mg, since there
is no force pointing in the neg direction away from the center;
see #120 AND #42, Ch. 5 , point B at top.

Set N = 0 IN THIS CASE.

Solve equations I and II simultaneously for h in terms of R
by eliminating v. The mass m cancels out.

(b) FOR THIS INITIAL HEIGHT h, greater than that of part
(a), the object never loses contact with the track at the top so
it does reach point C.

(II) CIRCULAR DYNAMICS: At C, the normal force points right
in the pos direction toward the center of the
circle. The weight of magnitude points vertically down and
does not contribute to the centripetal force. However, the weight
does contribute to the tangential acceleration since
the weight at C is tangent to the circle. Write:
(II) For the centripetal (center seeking) force, sum of forces in
radial direction = m*v2/R = pos - neg
= N - 0 = N, since there is no force pointing in the
neg direction away from the center. Note, at C the radial
acceleration is v2/R . You cannot find N without
the mass, not given.

(III) For the tangential acceleration, let the pos direction
be vertically down, a direction tangent to the circle at point
C.
Sum of forces in tangential direction = m*at
= mg - 0 = mg, where at is the tangential
acceleration; m cancels out.
DRAW RADIAL AND TANGENTIAL acceleration components to scale.

47.* ANOTHER CLASSIC: m*g*H = TOTAL HEAT, SINCE THE INITIAL
AND FINAL kinetic energies are zero. The Heat is the total heat
after possibly numerous trips across the flat section. TOTAL HEAT
=
fk*L, where L is the total distance moved on flat
surface before coming to rest. To find the number of trips,
divide TOTAL HEAT by fk*D, where D =
30 m exactly. From this ratio you can find the number of
complete trips; from the fractional remainder, you can find
exactly where the object comes to rest on the flat surface. If the
ratio is less than one, then the object does not make a complete
trip across the flat surface.

GO BACK TO CHAPTER 3 , review horizontal launch problems and solve for the speed vf at
the top satisfying the given condition. Then find vi
using conservation of energy.

55.* KEi + Ui = KEf
+ Uf + HEAT , where HEAT = 0.

Let the zero of gravitational energy be the ground.

# Initially (i) the system is at REST and Ui
= the gravitational potential energy of the 12.0 kg clock at the
given initial height.

# Finally (f), the 12.0 kg block is about to hit the ground and
the 4.0 kg block is at the same height that the 12.0 kg had
initially. Uf = the potential energy of the 4.0
kg block which is at height of 2.00 m above the ground. Both
blocks are moving at the same speed vf
even though they are moving in the opposite directions---use
that information to write KEf as the sum
of the kinetic energies of the two blocks.

Solve for the final common speed vf
of the two blocks.

56.*
USE KEi + Ui + WF
= KEf + Uf + HEAT, where U is the
gravitational potential energy in this case, KE is the
kinetic energy, WF is the work done by
the external thrust force and the HEAT is fk*D. The
constant friction force of magnitude fk = 500
(N) is given; D is the distance moved along ramp.

# Initially (i), the rocket is at REST and Ui = the
gravitational energy at the initial vertical height H above the
ground. Note the height H has a simple relationship with D,
the distance moved along the ramp; use simple geometry, the angle
of 53 degrees, and the
properties of right triangles to write H in terms of
D.

WF = F*D, where D is the distance moved along the
ramp and F is the magnitude of the given external thrust force =
2000 (N) exactly.

63.* ANOTHER CLASSIC: THE STRUCTURE OF THIS PROBLEM,
LIKE 9, and 46, mixes ENERGY CONSERVATION (Ch. 7)
and
CIRCULAR DYNAMICS (Ch. 5)

ENERGY CONSERVATION (Ch. 7):

(I) KEi + Ui = KEf + Uf
+ HEAT , where HEAT = 0.

# Initially (i), skier is at the top and is essentially at
rest; KEi =
0. Ui = m*g*R if the zero of potential
energy is considered to be at the level of the snow ball's
horizontal diameter.
(I) mg*R = (1/2)*m*vf2 + mg*R*cos
alpha, where alpha is angle with vertical shown.
# Finally (f), we consider the point when the skier leaves the
snow ball , i.e. when the normal of magnitude N becomes
zero. See figure 7.63. We define the location of this
point in the following way: Draw a radial line from the snow ball
center to this point and define the angle with the vertical
as alpha . KEf = (1/2)*m*vf2
and Uf = m*g*R*cos alpha.

CIRCULAR DYNAMICS: At any point on the surface of the ball, the normal force points
radially away from the center of the
circle. The gravitational force of magnitude mg points vertically down and
the component of the gravitational force along the radial
line is mg*cos theta and points toward the center of the
circle. Write:
(II) For the centripetal (center seeking) force, Sum of forces in
radial direction =
m*v2/R = pos - neg
= mg*cos theta - N.

For this problem set N = 0. Thus:

m*vf2/R = mg*cos theta.

Solve equations I and II simultaneously for cos theta by eliminating v.
Compute theta by evaluating cos -1. Your angle will
be between 45 and 90 degrees.

68. WE DO THIS PROBLEM USING TWO STYLES--FROM CH. 6 AND
7.
See #50 above.

CH. 6 STYLE:
(i) (1/2)*m*vf2 - (1/2)*m*vi2
= total work = Wg + Wf + WN,
work by gravity, the friction force and normal force respectively
where last term is zero since the normal work on the
way up slope must be vanish because that force is perpendicular to the motion i.e.
makes an angle of 90 degrees. ALSO, THE SLOPE IS FRICTIONLESS
MAKING WORK OF FRICTION Wf = 0.

(ii) FIND THE SPEED AT THE TOP OF THE CLIFF USING
(1/2)*m*vf2 - (1/2)*m*vi2
= total work = Wg WHERE THE NET VERTICAL
DISPLACEMENT IS FROM LOW TO HIGH. IS GRAVITY WORK POSITIVE
OR NEGATIVE?

(ii) AFTER LEAVING CLIFF, DURING FREE FALL (i.e., PROJECTILE
MOTION) THE gravitational work will be
positive for this high to low motion. IN THAT PHASE OF
PROBLEM YOU CAN USE CH. 3 METHODS AS IN LAB 3.

HERE'S HOW YOU DO IT CH. 7 STYLE:
FIND THE SPEED AT THE TOP OF THE CLIFF USING (1/2)*m*vi2
= (1/2)*m*vf2 + mgh. THEN USE
CH. 3 METHODS (PROJECTILE MOTION.)

70 (a) KEi + Ui = KEf + Uf
+ HEAT , where HEAT = 0. The block is released from rest
so that should tell you the value of KEi . The maximum
speed occurs when U is a minimum and that location is shown in
figure 7.42. Uf = 0 assuming both springs are
un-deformed and the block is in equilibrium. Ui =
sum of two spring potential energies: Spring 1 is stretched by
0.15 m and spring 2 is compressed by 0.15 m. To evaluate Ui,
apply the formula (1/2)*k*x2 for the
potential energy to both springs and add the two results.
Solve for KEf and the maximum speed vf
of the block WHICH OCCURS WHEN BOTH STRINGS ARE
UN-DEFORMED.
(b) The simplest way to do this part is to assume the
initial position (i) of the block is the same as in part (a) but
the final location (f) is when Spring 1 is at maximum compression. When
Spring 1 is at maximum compression, the block is at rest
momentarily (i): Ui = Uf
= sum of two spring potential energies. Spring 1
is compressed a distance |x'| and Spring 2 is stretched by the same
|x'| . Ui was evaluated in part
(a). Plug the symbol |x'| into the formula for Uf
and solve for |x'| numerically.

75. * SELECTIVELY USE KEi + Ui
+ WF = KEf + Uf
where U is the potential energy due to the spring force in
this case, KE is the kinetic energy and WF
is the work done by the external force of magnitude F. NO
FRICTION SO HEAT = 0.
The FULL equation with WF only applies between
initial point A and point-B but we know the process
continues beyond that. The Block will continue to move
beyond point-B and eventually come to rest. At that point,
the full distance from the origin is the amplitude A of
subsequent simple harmonic motion about the origin and would be
described by the function Acoswt for t >0 where t=0 is when the
block reaches maximum distance from origin---see Ch.
14 and virtual lab 2.. So that is your task---to find that
distance . Here are some tips:
(a) Find the speed of the block at point-B using :
KEA + UA + WF = KEB
+ UB
where WF = F*D. Note KEA = 0 and UA
= 0. Also WF = F*D, D = 0.25 m and UB
= (1/2)*k*(0.25)2 in Joules. Find the final
kinetic energy and final speed at B.
(b) Find the additional distance before coming to rest using:
KEB + UB = KEC + UC
. NOTE: KEB is known from the previous
part and UB = (1/2)*k*(0.25)2
Also KEC = 0 and UC = (1/2)*k*xC2
. Find xC. Then find 0.60 m - xC
.Note: KEA + UA + WF
= KEC + UC or, WF =
KEC + UC, where WF = F*D,
D = 0.25 m and KEC = 0.